MathOlympiadQues&Sol

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Volume 9 , Number 4 Volume 9 , Number 4 Volume 9 , Number 4 Volume 9 , Number 4 October 2004 October 2004October 2004October 2004 D ecember 2004 December 2004December 2004December 2004

Homothety

Kin Y. Li

Olympiad CornerThe Czech-Slovak-Polish Match this

year took place in Bilovec on June

21-22, 2004.Here are the problems.

Problem 1. Show that real numbersp, q,

rsatisfy the condition

p4(q r)

2+ 2p

2(q + r) + 1 =p

4

if and only if the quadratic equations

x2 +px + q = 0 andy

2 py + r= 0

have real roots (not necessarily distinct)which can be labeled byx1,x2 andy1,y2,

respectively, in such way that the

equalityx1y1 x2y2 = 1 holds.

Problem 2. Show that for each natural

number k there exist at most finitely

many triples of mutually distinct primes

p, q, rfor which the number qr kis a

multiple ofp, the number pr k is a

multiple ofq, and the numberpq kis a

multiple ofr.

Problem 3. In the interior of a cyclicquadrilateral ABCD, a point P is given

such that |BPC|=|BAP|+|PDC|.

Denote by E, F and G the feet of the

perpendiculars from the point P to the

lines AB, AD and DC, respectively.

Show that the triangles FEG and PBC

are similar.(continued on page 4)

Editors: (CHEUNG Pak-Hong), Munsang College, HK (KO Tsz-Mei)

(LEUNG Tat-Wing)

(LI Kin-Yin), Dept. of Math., HKUST

(NG Keng-Po Roger), ITC, HKPU

Artist: (YEUNG Sau-Ying Camille), MFA, CU

Acknowledgment: Thanks to Elina Chiu, Math. Dept.,

HKUST for general assistance.

On-line:http://www.math.ust.hk/mathematical_excalibur/

The editors welcome contributions from all teachers and

students. With your submission, please include your name,

address, school, email, telephone and fax numbers (if

available). Electronic submissions, especially in MS Word,

are encouraged. The deadline for receiving material for the

next issue isJanuary 20, 2005.

For individual subscription for the next five issues for the

03-04 academic year, send us five stamped self-addressed

envelopes. Send all correspondence to:

Dr. Kin-Yin LI

Department of Mathematics

The Hong Kong University of Science and Technology

Clear Water Bay, Kowloon, Hong Kong

Fax: (852) 2358 1643

Email: [email protected]

A geometric transformation of theplane is a function that sends everypoint on the plane to a point in the sameplane. Here we will like to discuss onetype of geometric transformations,called homothety, which can be used tosolve quite a few geometry problems insome international math competitions.

A homothety with center O and ratio kis a function that sends every point X

on the plane to the pointX such that

' .OX k OX =uu uur uu ur

So if |k| > 1, then the homothety is amagnification with center O. If |k| < 1,it is a reduction with center O. Ahomothety sends a figure to a similar

figure. For instance, let D, E, Fbe themidpoints of sides BC, CA, ABrespectively ofABC. The homothetywith center A and ratio 2 sends AFE

to ABC. The homothety with center atthe centroid G and ratio 1/2 sendsABCto DEF.

Example 1. (1978 IMO) In ABC, AB= AC. A circle is tangent internally tothe circumcircle ofABCand also to thesides AB, AC at P, Q, respectively.Prove that the midpoint of segment PQis the center of the incircle ofABC.

D

A

B' C'

O

P QI

B C

Solution. Let O be the center of the

circle. Let the circle be tangent to the

circumcircle ofABCatD. LetIbe the

midpoint of PQ. Then A, I, O, D are

collinear by symmetry. Consider the

homothety with center A that sends

ABC to ABC such that D is on

BC. Thus, k=AB/AB. As right

triangles AIP, ADB, ABD, APO are

similar, we have

AI /AO = (AI / AP)(AP / AO)

= (AD /AB)(AB /AD) =AB/AB=1/k.

Hence the homothety sends I to O.

Then O being the incenter ofABC

impliesIis the incenter ofABC.

Example 2. (1981 IMO) Three

congruent circles have a common point

O and lie inside a given triangle. Each

circle touches a pair of sides of the

triangle. Prove that the incenter and thecircumcenter of the triangle and the

point O are collinear.A

B C

A'

B'

O

C'

Solution. Consider the figure shown.Let A, B, C be the centers of thecircles. Since the radii are the same, so

AB is parallel toAB,BC is parallel toBC, CA is parallel to CA. Since AA,BB CC bisect A, B, Crespectively, they concur at the incenter

IofABC. Note O is the circumcenterofABC as it is equidistant from A,

B, C. Then the homothety with centerIsending ABC to ABCwill send Oto the circumcenter P of ABC.Therefore,I, O, P are collinear.

Example 3. (1982 IMO) A

non-isosceles triangle A1A2A3 is givenwith sides a1, a2, a3 (ai is the side

oppositeAi). For all i=1, 2, 3, Mi is the

midpoint of side ai, and Ti is the point

where the incircle touchs side ai. Denote

by Si the reflection ofTi in the interior

bisector of angleAi.

Prove that the linesM1S1,M2S2 andM3S3

are concurrent.

A2

A3 A1

I

T2

T1

T3

B2

B3

B1 S3

S2

S1


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Math ematical ExcaliburM athematical ExcaliburMath ematical ExcaliburM athem atical E xcalibur, Vol. 9, No. 4, Oct. 04- Dec. 04 Page 2

Solution. Let I be the incenter of

A1A2A3. Let B1, B2, B3 be the points

where the internal angle bisectors of

A1, A2, A3 meet a1, a2, a3

respectively. We will show SiSj is

parallel toMiMj. With respect toA1B1,

the reflection of T1 is S1 and the

reflection ofT2 is T3. So T3IS1 = T2IT1. With respect to A2B2, the

reflection ofT2 is S2 and the reflection

ofT1 is S3. SoT3IS2 =T1IT2. Then

T3IS1 = T3IS2. Since IT3 is

perpendicular to A1A2, we get S2S1 is

parallel to A1A2. SinceA1A2 is parallel

to M2M1, we get S2S1 is parallel to

M2M1. Similarly, S3S2 is parallel to

M3M2 and S1S3 is parallel toM1M3.

Now the circumcircle ofS1S2S3 is the

incircle of A1A2A3 and the

circumcircle ofM1M2M3 is the nine

point circle ofA1A2A3. Since A1A2A3

is not equilateral, these circles have

different radii. Hence S1S2S3 is not

congruent to M1M2M3 and there is a

homothety sending S1S2S3 to

M1M2M3. ThenM1S1,M2S2 andM3S3

concur at the center of the homothety.

Example 4. (1983 IMO) LetA be one

of the two distinct points ofintersection of two unequal coplanar

circles C1 and C2 with centers O1 and

O2 respectively. One of the common

tangents to the circles touches C1 at P1

and C2 at P2, while the other touches C1

at Q1 and C2 at Q2. Let M1 be the

midpoint of P1Q1 and M2 be the

midpoint ofP2Q2. Prove thatO1AO2

=M1AM2.

C2

C1O2

Q2

O1

A

O

P2

P1

Q1

M2

M1

B

Solution. By symmetry, lines O2O1,

P2P1, Q2Q1 concur at a point O.

Consider the homothety with center O

which sends C1 to C2. Let OA meet C1

atB, thenA is the image ofB under the

homothety. Since BM1O1 is sent toAM2O2, so M1BO1 =M2AO2.

Now OP1O1 similar to OM1P1

implies OO1/OP1 = OP1/OM1. Then

OO1 OM1 = OP12

= OA OB,

which implies points A, B, M1, O1 are

concyclic. Then M1BO1 = M1AO1.

Hence M1AO1 = M2AO2. Adding

O1AM2 to both sides, we haveO1AO2

=M1AM2.

Example 5. (1992 IMO) In the plane let C

be a circle,L a line tangent to the circle C,

andMa point onL. Find the locus of all

points P with the following property:

there exist two points Q,R onL such that

M is the midpoint of QR and C is the

inscribed circle ofPQR.

T C

C'L

P

Q RS

T'

U

M

V

Solution. Let L be the tangent to Cat S.

Let Tbe the reflection ofSwith respect to

M. Let Ube the point on Cdiametrically

opposite S. Take a point P on the locus.

The homothety with center P that sends C

to the excircle C will send U to T, the

point where QR touches C. Let line PR

touch C at V. Let s be the semiperimeter

ofPQR, then

TR = QS = s PR = PV PR =VR = TR

so that P, U, T are collinear. Then the

locus is on the part of line UT, opposite

the ray UTu uur

.

Conversely, for any point P on the part of

line UT, opposite the ray U Tu u ur

, the

homothety sends Uto Tand T, so T= T.

Then QS= s PR = PV PR =VR = TR =

TR and QM= QSMS=TR MT=RM.

Therefore, P is on the locus.For the next example, the solution

involves the concepts of power of a point

with respect to a circle and the radical axis.

We will refer the reader to the article

Power of Points Respect to Circles,

inMath Excalibur, vol. 4, no. 3,pp. 2,

4.

Example 6. (1999 IMO) Two circles 1

and 2 are inside the circle , and are

tangent to at the distinct pointsMand

N, respectively. 1 passes through thecenter of2. The line passing through

the two points of intersection of1 and

2 meets at A and B. The lines MA

and MB meet 1 at C and D,

respectively. Prove that CD is tangent

to 2.

L

A,A'

O1

C C'

O2

N

FE

BM

D

12

Solution. (Official Solution) LetEFbe

the chord of which is the common

tangent to 1 and 2 on the same side of

line O1O2 asA. LetEFtouch 1 at C.The homothety with center M that

sends1 to will send C to some point

A and lineEFto the tangent lineL of

atA. Since linesEFandL are parallel,

A must be the midpoint of arc FAE.

Then AEC = AFC = AME.

So AECis similar to AME. Then

the power ofA with respect to 1 is

AC AM= AE2. Similar, the power

ofA with respect to 2 is AF2. Since

AE=AF,A has the same power with

respect to 1 and 2. So A is on the

radical axis AB. Hence, A = A. Then

C = Cand Cis on EF.

Similarly, the other common tangent to

1 and 2 passes throughD. Let Oi be

the center of i. By symmetry with

respect to O1O2, we see that O2 is the

midpoint of arc CO2D. Then

DCO2 =CDO2 =FCO2.

This implies O2 is on the angle bisector

ofFCD. Since CF is tangent to 2,

therefore CD is tangent to 2.

(continued on page 4)


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Problem Corner

We welcome readers to submit their

solutions to the problems posed below

for publication consideration. The

solutions should be preceded by the

solvers name, home (or email) address

and school affiliation. Please send

submissions to Dr. Kin Y. Li,Department of Mathematics, The Hong

Kong University of Science &

Technology, Clear Water Bay, Kowloon,

Hong Kong. The deadline for

submitting solutions is January 20,

2005.

Problem 211. For every a, b, c, d in

[1,2], prove that

4 .a b c d a c

b c d a b d

+ + ++

+ + +

(Source: 32nd

Ukranian Math

Olympiad)

Problem 212. Find the largestpositive

integerNsuch that ifSis any set of 21

points on a circle C, then there existN

arcs ofCwhose endpoints lie in Sand

each of the arcs has measure not

exceeding 120.

Problem 213. Prove that the set of all

positive integers can be partitioned into

100 nonempty subsets such that if threepositive integers a, b, c satisfy a + 99 b

= c, then at least two of them belong to

the same subset.

Problem 214. Let the inscribed circle

of triangleABCbe tangent to sidesAB,

BC at E and F respectively. Let the

angle bisector of CAB intersect

segmentEFat K. Prove thatCKA is

a right angle.

Problem 215. Given a 88 board.

Determine all squares such that if eachone is removed, then the remaining 63

squares can be covered by 21 31

rectangles.

*****************

Solutions

****************

Problem 206. (Due to Zdravko F.

Starc, Vrac, Serbia and Montenegro)

Prove that ifa, b are the legs and c is

the hypotenuse of a right triangle, then

( ) ( ) 2 2 .a b a a b b c c+ + <

Solution. Cheng HAO (The SecondHigh School Attached to Beijing

Normal University), HUI Jack (QueensCollege, Form 5), D. Kipp JOHNSON(Valley Catholic School, Teacher,Beaverton, Oregon, USA), POON MingFung(STFA Leung Kau Kui College,Form 7), Achilleas P. PORFYRIADIS(American College of Thessaloniki Anatolia,Thessaloniki, Greece), Problem GroupDiscussion Euler-Teorema(Fortaleza, Brazil),Anna Ying PUN (STFA Leung Kau Kui

College, Form 6), TO Ping Leung (St. PetersSecondary School) and YIM Wing Yin(South Tuen Mun Government SecondarySchool, Form 4).

By Pythagoras theorem,

2 2( ) ( ) 2 .a b a b a b c+ + + =

Equality if and only if a = b. By the

Cauchy-Schwarz inequality,

( ) ( )a b a a b b+ +

2 2

( ) ( )a b a b a b + + +

2 2 .c c

For equality to hold throughout, we need

: : 1 :1a b a b a b+ = = , which

is not possible for legs of a triangle. So

we must have strict inequality.

Other commended solvers: HUDREA

Mihail (High School Tiberiu Popoviciu

Cluj-Napoca Romania) and TONG Yiu

Wai (Queen Elizabeth School, Form 7).

Problem 207. LetA = { 0, 1, 2,

, 9} andB1, B2, , Bk be nonempty subsets ofA

such thatBi andBj have at most 2 common

elements whenever i j. Find the

maximum possible value ofk.

Solution. ChengHAO (The Second HighSchool Attached to Beijing NormalUniversity), HUI Jack (Queens College,Form 5), POON Ming Fung(STFALeung Kau Kui College, Form 7) andAchilleas P. PORFYRIADIS (AmericanCollege of Thessaloniki Anatolia,Thessaloniki, Greece).

If we take all subsets ofA with 1, 2 or 3

elements, then these 10 + 45 + 120 = 175

subsets satisfy the condition. So k 175.

Let B1, B2, , Bk satisfying the condition

with kmaximum. If there exists aBi with

at least 4 elements, then every 3 element

subset ofBi cannot be one of theBj,ji,

sinceBi andBj can have at most 2 common

elements. So adding these 3 element

subsets toB1,B2, ,Bkwill still satisfy the

conditions. Since Bi has at least four 3element subsets, this will increase k,

which contradicts maximality of k. Then

every Bi has at most 3 elements. Hence, k

175. Therefore, the maximum kis 175.

Other commended solvers: CHANWai Hung (Carmel Divine GraceFoundation Secondary School, Form6), LI Sai Ki (Carmel Divine GraceFoundation Secondary School, Form6), LING Shu Dung,Anna Ying PUN(STFA Leung Kau Kui College, Form 6)and YIM WingYin (South Tuen MunGovernment Secondary School, Form4).

Problem 208. In ABC,AB >AC>BC.

LetD be a point on the minor arcBCof

the circumcircle ofABC.Let O be the

circumcenter ofABC. LetE, Fbe the

intersection points of lineAD with the

perpendiculars from O to AB, AC,

respectively. Let P be the intersection

of linesBEand CF. IfPB = PC+ PO,

then findBACwith proof.

Solution. Achilleas P. PORFYRIADIS

(American College of ThessalonikiAnatolia, Thessaloniki, Greece),Problem Group Discussion Euler -Teorema ( Fortaleza, Brazil) and AnnaYing PUN (STFA Leung Kau Kui College,Form 6).

O

D

A

B C

E

F

P

SinceEis on the perpendicular bisector

of chord AB and F is on the

perpendicular bisector of chordAC,AE

= BE and AF = CF. Applying exterior

angle theorem,

BPC=AEP +CFD

= 2 (BAD+CAD)= 2BAC=BOC.

Hence, B, C, P, O are concyclic. By

Ptolemys theorem,

PBOC= PCOB + POBC.

Then (PB PC)OC = POBC. Since

PB PC= PO, we get OC=BCand so

OBCis equilateral. Then

BAC= 12BOC= 30

Other commended solvers: Cheng HAO(The Second High School Attached toBeijing Normal University), HUI Jack(Queens College, Form 5), POONMing Fung(STFA Leung Kau KuiCollege, Form 7), TONG Yiu Wai


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(Queen Elizabeth School, Form 7) andYIM Wing Yin (South Tuen MunGovernment Secondary School, Form4).

Problem 209. Prove that there are

infinitely many positive integers n such

that 2n + 2 is divisible by n and 2n + 1 is

divisible by n 1.Solution. D. Kipp JOHNSON (ValleyCatholic School, Teacher, Beaverton,Oregon, USA), POON MingFung(STFA Leung Kau Kui College,Form 7) and Problem Group DiscussionEuler-Teorema(Fortaleza, Brazil).

As 22 + 2 = 6 is divisible by 2 and 22 + 1

= 5 is divisible by 1, n = 2 is one such

number.

Next, suppose 2n + 2 is divisible by n

and 2n + 1 is divisible by n 1. We will

proveN= 2n+ 2 is another such number.

SinceN 1 = 2n + 1= (n 1)kis odd, so k

is odd and n is even. SinceN= 2n + 2 =

2(2n1 + 1) = nm and n is even, so m must

be odd. Recall the factorization

xi + 1 = (x + 1)(x

i1 xi3 + + 1)

for odd positive integer i. Since kis odd,

2N+ 2 = 2(2N1

+ 1) = 2(2(n1)k+ 1) is

divisible by 2(2n1

+ 1) = 2n + 2 = N

using the factorization above. Since m is

odd, 2N

+ 1 = 2nm

+ 1 is divisible by

2n

+ 1 = N 1. Hence, Nis also such anumber. As N > n, there will be

infinitely many such numbers.

Problem 210. Let a1 = 1 and

1

1

2

nn

n

aa

a+

= +

for n = 1, 2, 3, . Prove that for every

integer n > 1,

2

2

2na

is an integer.

Solution. G.R.A. 20 Problem Group(Roma, Italy), HUDREA Mihail (HighSchool Tiberiu Popoviciu Cluj-Napoca Romania), Problem GroupDiscussion Euler Teorema (Fortaleza,Brazil), TO Ping Leung (St. PetersSecondary School) and YIM Wing Yin(South Tuen Mun GovernmentSecondary School, Form 4).

Note an=pn/qn, wherep1 = q1 = 1,pn+1 =

pn2 + 2qn

2 , qn+1 = 2pnqn for n = 1, 2, 3, .

Then

2 2 2

22.

2 2

n

n n n

q

a p q=

It suffices to show by mathematical

induction thatpn2

2qn2

= 1 for n > 1. We

havep22

2q22

= 32

222

= 1. Assuming

case n is true, we get

pn+12

2qn+12

= (pn2

+ 2qn2)

22(2pnqn)

= (pn2 2qn

2)2 = 1.

Other commended solvers: Ellen CHANOn Ting (True Light Girls College, Form5), Cheng HAO (The Second HighSchool Attached to Beijing NormalUniversity), HUI Jack (Queens College,Form 5), D. Kipp JOHNSON (ValleyCatholic School, Teacher, Beaverton,Oregon, USA), LAW Yau Pui (CarmelDivine Grace Foundation SecondarySchool, Form 6), Asger OLESEN(Toender Gymnasium (grammar school),Denmark), POON Ming Fung(STFALeung Kau Kui College, Form 7),Achilleas P. PORFYRIADIS (AmericanCollege of Thessaloniki Anatolia,Thessaloniki, Greece), Anna Ying PUN(STFA Leung Kau Kui College, Form 6),

Steve ROFFE, TONG Yiu Wai (QueenElizabeth School, Form 7) and YEUNGWai Kit (STFA Leung Kau Kui College,Form 4).

Olympiad Corner

(continued from page 1)

Problem 4. Solve the system of equations

1

1,

x

xy z= +

1

1,

y

yz x= +

1

1

z

zx y= +

in the domain of real numbers.

Problem 5. In the interiors of the sidesAB,

BCand CA of a given triangleABC, points

K, L and M, respectively, are given such

that

| | | | | |.

| | | | | |

AK BL C M

KB LC M A= =

Show that the triangles ABC and KLM

have a common orthocenter if and only if

the triangleABCis equilateral.

Problem 6. On the table there are kheaps

of 1, 2, , k stones, where k3. In the

first step, we choose any three of the heaps

on the table, merge them into a single new

heap, and remove 1 stone (throw it away

from the table) from this new heap. In the

second step, we again merge some three of

the heaps together into a single new heap,

and then remove 2 stones from this new

heap. In general, in the i-th step wechoose any three of the heaps, which

contain more than i stones when

combined, we merge them into a single

new heap, then remove i stones from this

new heap. Assume that after a number of

steps, there is a single heap left on the

table, containing p stones. Show that

the numberp is a perfect square if and

only if the numbers 2k+2 and 3k+1 are

perfect squares. Further, find the least

number k for which p is a perfect

square.

Homothety(continued from page 2)

Example 7. (2000APMO) LetABCbe

a triangle. LetMandNbe the points in

which the median and the angle

bisector, respectively atA meet the side

BC. Let Q and P be the points in which

the perpendicular atNtoNA meetsMA

andBA respectively and O the point in

which the perpendicular at P to BA

meetsANproduced.

Prove that QO is perpendicular toBC.

A

BCM N

P

O

B' C'K,Q

Solution (due toBobby Poon). The

caseAB =ACis clear.

Without loss of generality, we may

assume AB >AC. LetANintersect the

circumcircle ofABCatD. Then

DBC=DAC = 12BAC

=DAB =DCB.

SoDB =DCandMD is perpendicular

toBC.

Consider the homothety with center A

that sends DBC to OBC. Then

OB = OC andBCis parallel toBC.

LetBC intersect PNat K. Then

OBK=DBC =DAB

= 90 AOP =OPK.

So points P, B, O, K are concyclic.

Hence BKO =BPO = 90 andBK = CK. Since BC || BC, this

implies K is on MA. Hence, K = Q.

Now BKO = 90 implies QO=KO

BC. Finally,BC||BC implies QO

is perpendicular toBC.

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MathOlympiadQues&Sol